Transformations of incidence geometries, permutations, examples, and the group structure of automorphisms.

1. Transformations

For this lecture, we will use the following convention: lines are subsets of the set of points. Thus, given a model

\[ \mathcal M=(\mathcal P,\mathcal L,I), \]

we assume that every line \(\ell \in \mathcal L\) is a subset \(\ell \subseteq \mathcal P\), and incidence is the usual relation of belonging:

\[ p\,I\,\ell \qquad \Longleftrightarrow \qquad p\in \ell . \]

This is not very restrictive for the incidence geometries we are studying. If \(\mathcal M=(\mathcal P,\mathcal L,\mathcal R)\) satisfies axioms I1 and I2, then each line \(\ell \in \mathcal L\) determines the set

\[ R(\ell ):=\{p\in \mathcal P:p\,\mathcal R\,\ell \}. \]

One can replace \(\mathcal L\) by the set of subsets \(\{R(\ell ):\ell \in \mathcal L\}\). This gives an isomorphic model in which lines really are sets of points.

So a transformation does two things:

  • it is a bijection from the set of points to itself;
  • it sends lines to lines.

We denote by \(\operatorname {Aut}(\mathcal M)\) the set of all transformations of \(\mathcal M\).

2. The one-point model

Consider the model with one point

\[ \mathcal P=\{\ast \} \]

and no lines.

There is only one function \(\mathcal P\to \mathcal P\), namely the function sending \(\ast \) to itself. This function is bijective. Since there are no lines, there is nothing to check about sending lines to lines. Therefore this model has exactly one transformation.

3. The two-point model

Consider the model \(\mathcal M_2\) with points

\[ \mathcal P=\{0,1\} \]

and one line

\[ \ell =\{0,1\}. \]

The identity map is a transformation. Indeed, it is a bijection, and

\[ \operatorname {id}(\ell )=\operatorname {id}(\{0,1\}) = \{\operatorname {id}(0),\operatorname {id}(1)\} = \{0,1\} = \ell . \]

Thus the identity sends the only line to a line.

There is also the map

\[ \begin{array}{rcl} r:\mathcal P &\longrightarrow& \mathcal P\\ 0 &\longmapsto& 1\\ 1 &\longmapsto& 0. \end{array} \]

This map is a bijection. Moreover,

\[ r(\ell )=r(\{0,1\})=\{r(0),r(1)\}=\{1,0\}=\{0,1\}=\ell . \]

Therefore \(r\) is a transformation.

There are no other transformations. Indeed, let \(f:\mathcal P\to \mathcal P\) be a transformation. Then \(f\) is a bijection, hence injective. If \(f(0)=0\), then \(f(1)\neq 0\), so \(f(1)=1\), and \(f=\operatorname {id}\). If \(f(0)=1\), then \(f(1)\neq 1\), so \(f(1)=0\), and \(f=r\). Therefore

\[ \operatorname {Aut}(\mathcal M_2)=\{\operatorname {id},r\}. \]

4. Permutations

For example, if \(\Sigma =\{0,1,2\}\), then the following are permutations:

\[ \begin{array}{rcl} f:\Sigma &\longrightarrow& \Sigma \\ 0 &\longmapsto& 0\\ 1 &\longmapsto& 1\\ 2 &\longmapsto& 2 \end{array} \qquad \begin{array}{rcl} g:\Sigma &\longrightarrow& \Sigma \\ 0 &\longmapsto& 1\\ 1 &\longmapsto& 0\\ 2 &\longmapsto& 2 \end{array} \qquad \begin{array}{rcl} h:\Sigma &\longrightarrow& \Sigma \\ 0 &\longmapsto& 1\\ 1 &\longmapsto& 2\\ 2 &\longmapsto& 0. \end{array} \]

The set \(\{0,1,2\}\) has \(6\) permutations. Indeed, to define a permutation, we have \(3\) choices for the image of \(0\). Once this is chosen, we have \(2\) choices left for the image of \(1\), because a permutation is injective. Then there is only \(1\) choice left for the image of \(2\). Thus there are \(3\cdot 2\cdot 1=6\) permutations.

More generally, if \(\Sigma \) has \(n\) elements, then

\[ |\operatorname {Perm}(\Sigma )|=n!. \]

Indeed, there are \(n\) choices for the image of the first element, \(n-1\) choices for the image of the second element, \(n-2\) choices for the image of the third element, and so on.

5. The smallest incidence geometry with three points

Let \(\mathcal M_3\) be the model with points

\[ \mathcal P=\{0,1,2\} \]

and lines

\[ \{0,1\},\qquad \{0,2\},\qquad \{1,2\}. \]

Thus the lines are exactly the subsets of \(\mathcal P\) containing two elements.

We claim that

\[ \operatorname {Aut}(\mathcal M_3)=\operatorname {Perm}(\{0,1,2\}). \]

First, every transformation is a permutation of \(\{0,1,2\}\), because a transformation is, by definition, a bijection from the point set to itself. Thus

\[ \operatorname {Aut}(\mathcal M_3)\subseteq \operatorname {Perm}(\{0,1,2\}). \]

Conversely, let \(f\in \operatorname {Perm}(\{0,1,2\})\). We show that \(f\) is a transformation. Let \(\ell \) be a line. Since lines in \(\mathcal M_3\) are exactly the two-point subsets, we can write

\[ \ell =\{a,b\} \]

with \(a\neq b\). Since \(f\) is injective, \(f(a)\neq f(b)\). Hence

\[ f(\ell )=f(\{a,b\})=\{f(a),f(b)\} \]

is a two-point subset of \(\{0,1,2\}\). Therefore \(f(\ell )\) is a line. Hence \(f\) sends lines to lines, and so \(f\) is a transformation.

Thus

\[ \operatorname {Aut}(\mathcal M_3)=\operatorname {Perm}(\{0,1,2\}). \]

In particular,

\[ |\operatorname {Aut}(\mathcal M_3)|=3!=6. \]

6. The affine plane of order \(2\)

The affine plane of order \(2\) has \(4\) points. We label them

\[ \mathcal P=\{0,1,2,3\}. \]

In this model, every line contains exactly \(2\) points, and any two distinct points determine a unique line. Therefore the lines are exactly all two-point subsets of \(\mathcal P\).

We claim that

\[ \operatorname {Aut}(\mathcal M)=\operatorname {Perm}(\{0,1,2,3\}). \]

Again, every transformation is a permutation of the point set. Conversely, let \(f\in \operatorname {Perm}(\{0,1,2,3\})\). Let \(\ell \) be a line. Since every line is a two-point subset, we can write \(\ell =\{a,b\}\), with \(a\neq b\). Since \(f\) is injective, \(f(a)\neq f(b)\). Therefore

\[ f(\ell )=\{f(a),f(b)\} \]

is a two-point subset of \(\mathcal P\). Since every two-point subset is a line in the affine plane of order \(2\), \(f(\ell )\) is a line. Therefore \(f\) is a transformation.

Thus

\[ \operatorname {Aut}(\mathcal M)=\operatorname {Perm}(\{0,1,2,3\}). \]

In particular, this model has

\[ 4!=24 \]

transformations.

7. Not every permutation is a transformation

In the previous two examples, every permutation of the point set was a transformation. This is not always true.

Consider the model with points

\[ \mathcal P=\{0,1,2\} \]

and lines

\[ \ell =\{0,2\}, \qquad m=\{1,2\}. \]

Define \(f:\mathcal P\to \mathcal P\) by

\[ \begin{array}{rcl} f:\mathcal P &\longrightarrow& \mathcal P\\ 0 &\longmapsto& 1\\ 1 &\longmapsto& 2\\ 2 &\longmapsto& 0. \end{array} \]

This is a permutation of \(\mathcal P\), but it is not a transformation. Indeed,

\[ f(\ell )=f(\{0,2\})=\{f(0),f(2)\}=\{1,0\}=\{0,1\}. \]

But \(\{0,1\}\) is not a line in this model. Therefore \(f\) does not send lines to lines.

This example shows that being a permutation is not enough. To be a transformation, a bijection must also preserve the incidence structure.

8. A five-point example

Consider the model with points

\[ \mathcal P=\{0,1,2,3,4\}, \]

and lines

\[ \{0,1\},\quad \{2,3,4\},\quad \{0,2\},\quad \{0,3\},\quad \{0,4\},\quad \{1,2\},\quad \{1,3\},\quad \{1,4\}. \]

This is the model appearing in Homework 7.

The line \(\{2,3,4\}\) is special because it is the only line containing three points. Let \(F\) be a transformation. Since \(F\) is bijective, the image of a three-point set is again a three-point set. Since \(F\) sends lines to lines, \(F(\{2,3,4\})\) must be a line with three points. But \(\{2,3,4\}\) is the only such line. Therefore

\[ F(\{2,3,4\})=\{2,3,4\}. \]

Since \(F\) is bijective, it sends complements to complements. The complement of \(\{2,3,4\}\) in \(\mathcal P\) is \(\{0,1\}\). Hence

\[ F(\{0,1\})=\{0,1\}. \]

This strongly restricts the possible transformations. In particular, \(F\) must permute the set \(\{2,3,4\}\), and it must also permute the set \(\{0,1\}\). This is the starting point for the Homework 7 problem.

9. Transformations of the real affine plane

The examples above were finite. We also know many transformations of the usual real affine plane \(\mathbb R^2\).

For example, translations are transformations. If \(v\in \mathbb R^2\), then

\[ T_v:\mathbb R^2\to \mathbb R^2, \qquad T_v(p)=p+v \]

is a bijection and sends lines to lines.

Rotations are also transformations. More generally, if \(A\in \operatorname {GL}_2(\mathbb R)\) and \(b\in \mathbb R^2\), then the map

\[ p\longmapsto Ap+b \]

is a transformation of the real affine plane. Later, we will study analogous transformations over \(\mathbb Q\).

10. Groups

The set of transformations of a model has more structure than just being a set. We can compose transformations.

11. Automorphisms form a group

We check the group properties.

First, we check closure. Let \(f,g\in \operatorname {Aut}(\mathcal M)\). Since \(f\) and \(g\) are bijections, their composition \(f\circ g\) is also a bijection. We now check that \(f\circ g\) sends lines to lines. Let \(\ell \in \mathcal L\). Since \(g\) is a transformation, \(g(\ell )\) is a line. Since \(f\) is a transformation, \(f(g(\ell ))\) is also a line. But

\[ f(g(\ell ))=(f\circ g)(\ell ). \]
Therefore \(f\circ g\) sends lines to lines, so \(f\circ g\in \operatorname {Aut}(\mathcal M)\).

The identity map is a transformation. It is bijective, and for every line \(\ell \), we have \(\operatorname {id}(\ell )=\ell \). Thus the identity map is the neutral element for composition.

Now let \(f\in \operatorname {Aut}(\mathcal M)\). Since \(f\) is a bijection, it has an inverse map \(f^{-1}:\mathcal P\to \mathcal P\). One must check that \(f^{-1}\) sends lines to lines. This is one of the homework exercises, and it uses axioms I1 and I2. Therefore \(f^{-1}\in \operatorname {Aut}(\mathcal M)\). By definition of inverse maps,

\[ f\circ f^{-1}=\operatorname {id} \qquad \text {and}\qquad f^{-1}\circ f=\operatorname {id}. \]

Finally, composition of maps is associative. Indeed, if \(f,g,h\in \operatorname {Aut}(\mathcal M)\), then for every point \(p\in \mathcal P\),

\[ ((f\circ g)\circ h)(p)=f(g(h(p)))=(f\circ (g\circ h))(p). \]
Hence
\[ (f\circ g)\circ h=f\circ (g\circ h). \]
Therefore \(\operatorname {Aut}(\mathcal M)\), with composition, is a group.